# Find $P(n+1)$ for a polynomial P

Fix a nonnegative integer $$n$$. Let $$P(x)$$ be a polynomial of degree $$n$$ (over the real numbers) such that for all $$k\in\left\{0,1,\ldots,n\right\}$$, we have $$P(k)=\dfrac{k}{k+1}$$. Find $$P(n+1)$$.

My attempt:

I consider a new polynomial $$Q(x)=(x+1)P(x)-x$$. This $$Q$$ is a polynomial of degree $$\leq n+1$$ satisfying: $$Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$$. Therefore $$Q(x)=\lambda x(x-1)\cdots(x-n)$$ for a certain real $$\lambda$$. Hence, $$Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$$, but also $$Q(-1)=(-1+1)P(-1)+1=1$$. Comparing these, we get $$\lambda \cdot (-1)^{n+1} (n+1)! = 1$$, thus $$\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$$. Since $$Q(n+1)=\lambda(n+1)!$$ then $$Q(n+1)=(-1)^{n+1}$$, and from $$P(x) = \dfrac{Q(x)+x}{x+1}$$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $$P(n+1)=\frac{n+1}{n+2}$$ but the term $$(-1)^{n+1}$$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?

• Try a couple of examples for small $n$. This won't prove that you are right but it may improve your confidence. It might prove that you are wrong. – badjohn Apr 26 '19 at 16:15

Your solution is fine.$$Q(-1) = \lambda (-1)(-2)\ldots (-(n+1))=\lambda (-1)^{n+1}(n+1)!=1$$

$$\lambda = \frac{(-1)^{n+1}}{(n+1)!}$$

$$Q(x)=\frac{(-1)^{n+1}x(x-1)\ldots (x-n)}{(n+1)!}=(x+1)P(x)-x$$

Let $$x=n+1$$

$$\frac{(-1)^{n+1}(n+1)n\ldots 1}{(n+1)!}=(n+2)P(n+1)-(n+1)$$

Hence $$P(n+1) = \frac{n+1 + (-1)^{n+1}}{n+2}$$

Let's check with small cases to see if $$(-1)^{n+1}$$ should be there.

Let $$n=1$$, then we have $$P(0)=0$$ and $$P(1)=\frac12$$. Then $$P(x) = \frac{x}2$$.

We have $$P(2)= 1$$.

Notice that $$\frac{n+1}{n+2}<1$$ and hence $$P(n) = \frac{n+1}{n+2}$$ is certainly wrong.

• That is the same as mine. Do you know an alternative proof avoiding the use of this specific polynomial $Q$? Typo fixed, forgot the frac. – DINEDINE Apr 26 '19 at 16:22
• nope, your approach seems elegant enough. – Siong Thye Goh Apr 26 '19 at 16:24